Cho hàm số $f\left( x \right)$ có đạo hàm lên tục trên đoạn...

Cách giải:
Ta có $\int\limits_{0}^{3}{xf\left( x \right)dx}=\dfrac{531}{20}$
Đặt $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv=xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=f'\left( x \right)dx \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$
Khi đó ta có:
$\int\limits_{0}^{3}{xf\left( x \right)dx}=\dfrac{531}{20}\Leftrightarrow \dfrac{{{x}^{2}}}{2}f\left( x \right)\left| \begin{aligned}
& 3 \\
& 0 \\
\end{aligned} \right.-\dfrac{1}{2}\int\limits_{0}^{3}{{{x}^{2}}f'\left( x \right)dx}=\dfrac{531}{20}$
$\Leftrightarrow \dfrac{9}{2}f\left( 3 \right)-\dfrac{1}{2}\int\limits_{0}^{3}{{{x}^{2}}f'\left( x \right)dx}=\dfrac{531}{20}$
$\Leftrightarrow 63-\dfrac{1}{2}\int\limits_{0}^{3}{{{x}^{2}}f'\left( x \right)dx}=\dfrac{531}{20}$
$\Leftrightarrow \int\limits_{0}^{3}{{{x}^{2}}f'\left( x \right)dx}=\dfrac{729}{10}$
Xét $\int\limits_{0}^{3}{\left[ f'\left( x \right)+k{{x}^{2}} \right]dx}=0$
$\Leftrightarrow \int\limits_{0}^{3}{{{\left[ f'\left( x \right) \right]}^{2}}dx}+2k\int\limits_{0}^{3}{{{x}^{2}}f'\left( x \right)dx}+{{k}^{2}}\int\limits_{0}^{3}{{{x}^{4}}dx}=0$
$\Leftrightarrow \dfrac{2187}{20}+2k.\dfrac{729}{10}+{{k}^{2}}\dfrac{{{x}^{5}}}{5}\left| \begin{aligned}
& 3 \\
& 0 \\
\end{aligned} \right.=0$
$\Leftrightarrow \dfrac{243}{5}{{k}^{2}}+\dfrac{729}{5}k+\dfrac{2187}{20}=0$
$\Leftrightarrow k=-\dfrac{3}{2}$
Khi đó ta có: $\int\limits_{0}^{3}{{{\left[ f'\left( x \right)-\dfrac{3}{2}{{x}^{2}} \right]}^{2}}dx}=0\Leftrightarrow f'\left( x \right)=\dfrac{3}{2}{{x}^{2}}.$
$\Rightarrow f\left( x \right)=\int\limits_{{}}^{{}}{\dfrac{3}{2}{{x}^{2}}dx}=\dfrac{{{x}^{3}}}{2}+C.$
Lại có $f\left( 3 \right)=14\Rightarrow 14=\dfrac{27}{2}+C\Leftrightarrow C=\dfrac{1}{2}\Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{2}+\dfrac{1}{2}$
Vậy $\int\limits_{0}^{3}{\left[ f\left( x \right)-1 \right]dx}=\int\limits_{0}^{3}{\left( \dfrac{{{x}^{3}}}{2}+\dfrac{1}{2}-1 \right)dx}=\dfrac{69}{8}.$